Java 创建 JSONObject 时 org.json 未报告的异常
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org.json unreported exception while creating a JSONObject
提问by lastmjs
Can anyone help me understand what is going wrong?
任何人都可以帮助我了解出了什么问题吗？
unreported exception org.json.JSONException; must be caught or declared to be thrown
jsonObj = new JSONObject("{\"count\":3939,\"has_more\":true,\"map_location\":{\"lat\":0.60996950000000183,\"lon\":-27.568517000000003,\"panoramio_zoom\":16},\"photos\":[{\"height\":375,}]}"); //creates the JSON object from the jsonString, for parsing
^
1 error
1 错误
I'm using org.json, and I think I have everything set up correctly. I'm trying to create a JSONObject using the constructor in org.json that takes a source string, and I keep getting this exception. I'm not sure what is wrong with the string that I am sending in. Thanks
我正在使用 org.json，我想我已经正确设置了所有内容。我正在尝试使用 org.json 中采用源字符串的构造函数创建一个 JSONObject，但我不断收到此异常。我不确定我发送的字符串有什么问题。谢谢
采纳答案by Yagnesh Agola
Catch your Exception by creating try and catch
通过创建 try 和 catch 来捕获您的异常
try {
JSONObject jsonObj = new JSONObject("{\"count\":3939,\"has_more\":true,\"map_location\":{\"lat\":0.60996950000000183,\"lon\":-27.568517000000003,\"panoramio_zoom\":16},\"photos\":[{\"height\":375,}]}");
System.out.println(jsonObj);
} catch (JSONException e) {
//some exception handler code.
}
Or either thorws your exception to caller method.
或者要么将您的异常抛给调用者方法。
public String yourMethod(String jsonString) throws JSONException
回答by Jigar Joshi
constructor declares to throworg.json.JSONExceptionso you must handle it (catch & handle or rethrow to let caller handle it)
构造函数声明抛出org.json.JSONException所以你必须处理它（捕获并处理或重新抛出让调用者处理它）
